CBSE Notes Class 10 Maths Chapter 7 Coordinate Geometry

1.0Introduction to Coordinate Geometry

Coordinate Geometry, also known as Cartesian Geometry, is a study of geometry using coordinates. In this chapter, we focus on the Cartesian coordinate system and distance formula to solve algebraically geometric problems. This CBSE Notes for Class 10 Would give an added advantage to the students.

2.0Download CBSE Notes for Class 10 Maths Chapter 7: Coordinate Geometry - Free PDF!!

Hey there! Looking to ace Coordinate Geometry in your Class 10 Maths? These free CBSE notes for Chapter 7 will give you a solid understanding of the concepts.

3.0CBSE Class 10 Maths Chapter 7 Coordinate Geometry - Revision Notes

What is the Cartesian Coordinate System?

1. It is a two-dimensional Cartesian coordinate system in which the position of any point can be defined with the aid of two perpendicular lines, and these lines are called axes.

• The horizontal axis is named as the x-axis or Abscissa.
• The y-axis is the vertical axis or Ordinate. 

2. If the x-axis and the y-axis intersect at a point, such an intersection is called origin. The origin is denoted by the letter O with the coordinate (0, 0).
3. Any point in the coordinate plane is represented as P(x, y), where x denotes the distance from the origin along the x-axis and y is the distance from the origin along the y-axis.

The Distance Formula 

In maths, the distance formula is used to find the distance between two points on the cartesian system, for example A(x₁,y₁) and B(x₂,y₂). The distance is always positive. The formula to find the distance between these points: 

$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance is two points if any one point is the origin.  $AB=\sqrt{(x{)}^2+(y{)}^2}$

Example 1: Find the distance between P(2, 6) and Q(8, 3) using the distance formula. 

Solution:

$PQ=\sqrt{(8-2{)}^2+(3-6{)}^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}$

Example 2: Find the point on the y-axis which is equidistant from A(2, 4) and B(4, 8). 

Solution: The point is on the y-axis, hence the coordinates will be (0, y)

$\sqrt{(0-2{)}^2+(y-4{)}^2}=\sqrt{(0-4{)}^2+(y-8{)}^2}$

Squaring both sides,

4 + y² - 8y + 16 = 16 + y² - 16y + 64 

-8y + 16y = 64 - 4

8y = 60 

y = 7.5

Note: If the given point is on the x-axis, then the y-coordinate will be 0 and vice-versa. 

Section Formula 

Section formula is used to find the coordinates of a line segment. For example, in the given figure, C(x, y) is the point cutting line segments A(x₁, y₁) and B(x₂, y₂) in the ratio m and n.

$\mathbf{X}=\frac{m\times x_2+n\times x_1}{m+n}$

$\mathrm{y}=\frac{m\times y_2+n\times y_1}{m+n}$

Here, (x, y) are the coordinates of C. 

Midpoint Formula: 

When point C cuts the line segment AB in two equal parts, meaning m = n, then we can use another formula that is:

$\mathrm{X}=\frac{x_2+x_1}{2}$

$\mathrm{y}=\frac{y_2+y_1}{2}$

Example: Find the coordinates of a point R(x, y) which is cutting the line segment P(2, 4) and Q(4, 8) in the ratio of 2 and 4. 

Solution:

$X=\frac{2\times 4+4\times 2}{2+4}=\frac{8+8}{6}=\frac{8}{3}$

$y=\frac{2\times 8+4\times 4}{2+4}=\frac{16+16}{6}=\frac{16}{3}$

$\text{ Coordinates of }\mathrm{P}\left(\frac{8}{3},\frac{16}{3}\right)\text{. }$

Example: Find the ratio of the point Q that is on the y-axis cutting the line segment R(5, 7) and P(9, 3). 

Solution: Let the ratio be K : 1.

The point Q is on the y-axis; hence, Q(0, y)

Coordinates of Q 

$0=\frac{k\times 9+1\times 5}{K+1}$

k9+15 = 0 

K = 9/5

Hence, The ratio is 9 : 5 

4.0Key Features of CBSE Maths Notes for Class 10 Chapter 7

• The notes are well aligned with the latest curriculum suggested by CBSE. 
• Every concept is simplified and explained with simple and easy language, making these notes ideal for self-learning. 
• The step-by-step guide is provided with every concept of coordinate geometry to ensure better understanding.